Question

\[x^2 + {1 \over x^2} = 7 \Longrightarrow x^3 + {1 \over x^3} = ? \]

Answer 1

x^4 - 7x^2 + 1 = 0
Let u = x^2
u^2 - 7u + 1 = 0
what are the roots if any?

Answer 2

\[\large{(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2}\]
\[\large{(x+\frac{1}{x})^2=7+2=9}\]

Answer 3

x3+1/x3
=(a3+b3)=(a+b)(a2-ab+b2)

Answer 4

x^2 = 7 -1/x^2 multiply by x


x^3 = 7x - 1/x

Answer 5

from the above we get :
\[\large{x+\frac{1}{x}=3}\]

right @ParthKohli ?

Answer 6

just subs in the equation

Answer 7

Ah! \(x = 3 - {1 \over x}\)?

Answer 8

no dont do that way parth

Answer 9

\[\large x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x})\]

Answer 10

\[\large{(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)}\]
\[\large{(3)(7-1)}\]
this way ..

Answer 11

apply the formula of \(\large{x^3+\frac{1}{x^3}}\)

Answer 12

\[\large x^3 + \frac{1}{x^3} = 3^3 - 3(3)\]

Answer 13

Yup. Got it.

Answer 14

nice to hear !!
best of luck dude

Answer 15

Thank you! @mathslover @waterineyes @lgbasallote @higgs @Yahoo! @sai1234

Answer 16

x3+1/x3 = (x+1/x)(x2+1x2-1)
use (x+1/x)2=x2+1x2+2

(x+1/x)2=7+2=9
get (x+1/x) value and substitute in (x+1/x)(x2+1x2-1)